kirktrue commented on code in PR #17700:
URL: https://github.com/apache/kafka/pull/17700#discussion_r1932702957
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clients/src/test/java/org/apache/kafka/clients/consumer/internals/FetchRequestManagerTest.java:
##########
@@ -3437,21 +3443,330 @@ public void testPollWithRedundantCreateFetchRequests()
{
}
- private OffsetsForLeaderEpochResponse prepareOffsetsForLeaderEpochResponse(
- TopicPartition topicPartition,
- Errors error,
- int leaderEpoch,
- long endOffset
- ) {
- OffsetForLeaderEpochResponseData data = new
OffsetForLeaderEpochResponseData();
- data.topics().add(new OffsetForLeaderTopicResult()
- .setTopic(topicPartition.topic())
- .setPartitions(Collections.singletonList(new EpochEndOffset()
- .setPartition(topicPartition.partition())
- .setErrorCode(error.code())
- .setLeaderEpoch(leaderEpoch)
- .setEndOffset(endOffset))));
- return new OffsetsForLeaderEpochResponse(data);
+ /**
+ * This test makes several calls to {@link #sendFetches()}, and after
each, the buffered partitions are
+ * modified to either cause (or prevent) a fetch from being requested.
+ */
+ @Test
+ public void testFetchRequestWithBufferedPartitions() {
+ buildFetcher();
+
+ // The test requires that there are multiple nodes as the fetch
request logic is based in part off of the
+ // partition-to-node relationship.
+ int numNodes = 2;
+ Set<TopicPartition> partitions = Set.of(tp0, tp1, tp2, tp3);
+ assignFromUser(partitions, numNodes);
+
+ // Get all the nodes serving as the leader for these partitions, and
then split them into the separate
+ // nodes and set of partitions to make things easier to keep track of
later.
+ List<Node> nodes = nodesForPartitionLeaders(partitions);
+ Node node0 = nodes.get(0);
+ Node node1 = nodes.get(1);
+ List<TopicPartition> node0Partitions = partitionsForNode(node0,
partitions);
+ List<TopicPartition> node1Partitions = partitionsForNode(node1,
partitions);
+
+ // Seek each partition so that it becomes eligible to fetch.
+ partitions.forEach(tp -> subscriptions.seek(tp, 0));
+
+ // sendFetches() call #1 should issue requests to node 0 or node 1
since neither has buffered data.
+ assertEquals(2, sendFetches());
+ prepareFetchResponses(node0, node0Partitions, 0);
+ prepareFetchResponses(node1, node1Partitions, 0);
+ networkClientDelegate.poll(time.timer(0));
+ assertEquals(4, fetcher.fetchBuffer.bufferedPartitions().size());
+ collectSelectedPartition(node0Partitions.remove(0), partitions);
+ assertEquals(3, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // sendFetches() call #2 shouldn't issue requests to either node 0 or
node 1 since they both have buffered data.
+ assertEquals(0, sendFetches());
+ networkClientDelegate.poll(time.timer(0));
+ collectSelectedPartition(node1Partitions.remove(0), partitions);
+ assertEquals(2, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // sendFetches() call #3 shouldn't issue requests to either node 0 or
node 1 since they both have buffered data.
+ assertEquals(0, sendFetches());
+ networkClientDelegate.poll(time.timer(0));
+ collectSelectedPartition(node0Partitions.remove(0), partitions);
+ assertEquals(1, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // Node 0's partitions have all been collected, so validate that and
then reset the list of partitions
+ // from which to fetch data so the next pass should request can fetch
more data.
+ assertTrue(node0Partitions.isEmpty());
+ node0Partitions = partitionsForNode(node0, partitions);
+
+ // sendFetches() call #4 should issue a request to node 0 since its
buffered data was collected.
+ assertEquals(1, sendFetches());
+ prepareFetchResponses(node0, node0Partitions, 10);
+ networkClientDelegate.poll(time.timer(0));
+ collectSelectedPartition(node1Partitions.remove(0), partitions);
+ assertEquals(2, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // Node 1's partitions have likewise all been collected, so validate
that and reset.
+ assertTrue(node1Partitions.isEmpty());
+ node1Partitions = partitionsForNode(node1, partitions);
+
+ // sendFetches() call #5 should issue a request to node 1 since its
buffered data was collected.
+ assertEquals(1, sendFetches());
+ prepareFetchResponses(node1, node1Partitions, 10);
+ networkClientDelegate.poll(time.timer(0));
+ assertEquals(4, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // Collect all the records and make sure they include all the
partitions, and validate that there is no data
+ // remaining in the fetch buffer.
+ assertEquals(partitions, fetchRecords().keySet());
+ assertEquals(0, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // sendFetches() call #6 should issue a request to nodes 0 and 1 since
its buffered data was collected.
+ assertEquals(2, sendFetches());
+ prepareFetchResponses(node0, node0Partitions, 20);
+ prepareFetchResponses(node1, node1Partitions, 20);
+ networkClientDelegate.poll(time.timer(0));
+ assertEquals(4, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // Just for completeness, collect all the records and make sure they
include all the partitions, and validate
+ // that there is no data remaining in the fetch buffer.
+ assertEquals(partitions, fetchRecords().keySet());
+ assertEquals(0, fetcher.fetchBuffer.bufferedPartitions().size());
+ }
+
+ @Test
+ public void testFetchRequestWithBufferedPartitionNotAssigned() {
+ buildFetcher();
+
+ // The test requires that there are multiple nodes as the fetch
request logic is based in part off of the
+ // partition-to-node relationship.
+ int numNodes = 2;
+ Set<TopicPartition> partitions = Set.of(tp0, tp1, tp2, tp3);
+ assignFromUser(partitions, numNodes);
+
+ // Get all the nodes serving as the leader for these partitions, and
then split them into the separate
+ // nodes and set of partitions to make things easier to keep track of
later.
+ List<Node> nodes = nodesForPartitionLeaders(partitions);
+ Node node0 = nodes.get(0);
+ Node node1 = nodes.get(1);
+ List<TopicPartition> node0Partitions = partitionsForNode(node0,
partitions);
+ List<TopicPartition> node1Partitions = partitionsForNode(node1,
partitions);
+
+ // Seek each partition so that it becomes eligible to fetch.
+ partitions.forEach(tp -> subscriptions.seek(tp, 0));
+
+ // sendFetches() call #1 should issue requests to node 0 or node 1
since neither has buffered data.
+ assertEquals(2, sendFetches());
+ prepareFetchResponses(node0, node0Partitions, 0);
+ prepareFetchResponses(node1, node1Partitions, 0);
+ networkClientDelegate.poll(time.timer(0));
+
+ // Grab both partitions for node 0. The first partition will be
collected so that it doesn't have anything
+ // in the fetch buffer. The second node will be left in the buffer,
but will be unassigned in a bit.
+ TopicPartition node0Partition1 = node0Partitions.remove(0);
+ TopicPartition node0Partition2 = node0Partitions.remove(0);
+
+ assertEquals(4, fetcher.fetchBuffer.bufferedPartitions().size());
+ collectSelectedPartition(node0Partition1, partitions);
+ assertEquals(3, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // Change the set of assigned partitions to exclude the remaining
buffered partition for node 0, which means
+ // that partition is unassigned.
+ Set<TopicPartition> topicsWithoutUnassignedPartition = new
HashSet<>(partitions);
+ topicsWithoutUnassignedPartition.remove(node0Partition2);
+ subscriptions.assignFromUser(topicsWithoutUnassignedPartition);
+
+ // The collected partition should have a retrievable position, but the
unassigned position should throw
+ // an error when attempting to retrieve its position.
+ assertDoesNotThrow(() -> subscriptions.position(node0Partition1));
+ assertThrows(IllegalStateException.class, () ->
subscriptions.position(node0Partition2));
+
+ // sendFetches() call #2 should issue a request to node 0 because the
first partition in node 0 was collected
+ // (and its buffer removed) and the second partition for node 0 was
unassigned. As a result, there are now no
+ // *assigned* partitions for node 0 that are buffered.
+ assertEquals(1, sendFetches());
+ }
+
+ @Test
+ public void testFetchRequestWithBufferedPartitionMissingLeader() {
+ buildFetcher();
+
+ // The test requires that there are multiple nodes as the fetch
request logic is based in part off of the
+ // partition-to-node relationship.
+ int numNodes = 2;
+ Set<TopicPartition> partitions = Set.of(tp0, tp1, tp2, tp3);
+ assignFromUser(partitions, numNodes);
+
+ // Get all the nodes serving as the leader for these partitions, and
then pull out the separate
+ // nodes and set of partitions to make things easier to keep track of
later.
+ List<Node> nodes = nodesForPartitionLeaders(partitions);
+ Node node0 = nodes.get(0);
+ Node node1 = nodes.get(1);
+ List<TopicPartition> node0Partitions = partitionsForNode(node0,
partitions);
+ List<TopicPartition> node1Partitions = partitionsForNode(node1,
partitions);
+
+ // Seek each partition so that it becomes eligible to fetch.
+ partitions.forEach(tp -> subscriptions.seek(tp, 0));
+
+ // sendFetches() call #1 should issue requests to node 0 and node 1
since neither has buffered data.
+ assertEquals(2, sendFetches());
+ prepareFetchResponses(node0, node0Partitions, 0);
+ prepareFetchResponses(node1, node1Partitions, 0);
+ networkClientDelegate.poll(time.timer(0));
+
+ TopicPartition node0Partition1 = node0Partitions.remove(0);
+ TopicPartition node0Partition2 = node0Partitions.remove(0);
+
+ // Per the fetch response, data for both of node 0's partitions are in
the fetch buffer.
+
assertTrue(fetcher.fetchBuffer.bufferedPartitions().contains(node0Partition1));
+
assertTrue(fetcher.fetchBuffer.bufferedPartitions().contains(node0Partition2));
+
+ // Collect node 0's first partition (node0Partition1) which will
remove it from the fetch buffer.
+ collectSelectedPartition(node0Partition1, partitions);
+
+ // Node 0's first partition (node0Partition1) was collected so it's
not in the fetch buffer, but node 0's
+ // second partition (node0Partition2) remains in the fetch buffer.
+
assertFalse(fetcher.fetchBuffer.bufferedPartitions().contains(node0Partition1));
+
assertTrue(fetcher.fetchBuffer.bufferedPartitions().contains(node0Partition2));
+
+ // Overwrite node0Partition2's position with an empty leader, but
verify that it is still buffered. Having
+ // a leaderless, buffered partition is key to triggering the test case.
+ subscriptions.position(node0Partition2, new
SubscriptionState.FetchPosition(
+ 0,
+ Optional.empty(),
+ Metadata.LeaderAndEpoch.noLeaderOrEpoch()
+ ));
+
assertTrue(fetcher.fetchBuffer.bufferedPartitions().contains(node0Partition2));
+
+ // Validate the state of the collected partition (node0Partition1) and
leaderless partition (node0Partition2)
+ // before sending the fetch requests.
+ Optional<Node> node0Partition1Leader =
subscriptions.position(node0Partition1).currentLeader.leader;
+ Optional<Node> node0Partition2Leader =
subscriptions.position(node0Partition2).currentLeader.leader;
+ assertTrue(node0Partition1Leader.isPresent());
+ assertEquals(node0, node0Partition1Leader.get());
+ assertFalse(node0Partition2Leader.isPresent());
+
+ // sendFetches() call #2 should issue a fetch request to node 0
because it has no buffered partitions:
+ //
+ // - node0Partition1 was collected and thus not in the fetch buffer
+ // - node0Partition2, while still in the fetch buffer, is leaderless
+ //
+ // As a result, there are now effectively no buffered partitions for
which node 0 is the leader.
+ assertEquals(1, sendFetches());
+ }
+
+ @Test
+ public void testFetchRequestWithBufferedPartitionMissingPosition() {
+ buildFetcher();
+
+ // The test requires that there are multiple nodes as the fetch
request logic is based in part off of the
+ // partition-to-node relationship.
+ int numNodes = 2;
+ Set<TopicPartition> partitions = Set.of(tp0, tp1, tp2, tp3);
+ assignFromUser(partitions, numNodes);
+
+ // Get all the nodes serving as the leader for these partitions, and
then split them into the separate
+ // nodes and set of partitions to make things easier to keep track of
later.
+ List<Node> nodes = nodesForPartitionLeaders(partitions);
+ Node node0 = nodes.get(0);
+ Node node1 = nodes.get(1);
+ List<TopicPartition> node0Partitions = partitionsForNode(node0,
partitions);
+ List<TopicPartition> node1Partitions = partitionsForNode(node1,
partitions);
+
+ // Seek each partition so that it becomes eligible to fetch.
+ partitions.forEach(tp -> subscriptions.seek(tp, 0));
+
+ // sendFetches() call #1 should issue requests to node 0 or node 1
since neither has buffered data.
+ assertEquals(2, sendFetches());
+ prepareFetchResponses(node0, node0Partitions, 0);
+ prepareFetchResponses(node1, node1Partitions, 0);
+ networkClientDelegate.poll(time.timer(0));
+
+ // Grab both partitions for node 0. The first partition will be
collected so that it doesn't have anything
+ // in the fetch buffer. The second node will be left in the buffer,
but will clear out its position's leader
+ // node.
+ TopicPartition node0Partition1 = node0Partitions.remove(0);
+ TopicPartition node0Partition2 = node0Partitions.remove(0);
+
+ assertEquals(4, fetcher.fetchBuffer.bufferedPartitions().size());
+ collectSelectedPartition(node0Partition1, partitions);
+ assertEquals(3, fetcher.fetchBuffer.bufferedPartitions().size());
+
+ // Overwrite the position with an empty leader to trigger the test
case.
+ subscriptions.position(node0Partition2, null);
+
+ // Both the collected partition and the position without a partition
leader should have a retrievable position.
Review Comment:
This was changed to:
```java
// Confirm that calling SubscriptionState.position() succeeds for a
leaderless partition. While it shouldn't
// throw an exception, it should return a null position.
SubscriptionState.FetchPosition position = assertDoesNotThrow(() ->
subscriptions.position(node0Partition2));
assertNull(position);
```
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